Question

In the figure given above, the ratio $\frac{AreaofDPQC}{AreaofABCD}$ =____.

• A. 1
• B. $\frac{1}{2}$
• C. $\frac{2}{3}$
• D. $\frac{1}{3}$

Answer 1

The correct option is C $\frac{2}{3}$


Let h be the height between the parallel lines AB and CD

$PQ=\frac{1}{3}AB$

Hence, area of DPQC = $\frac{1}{2}\times htimes(PQ+DC)$

= $\frac{h}{2}\times (DC+\frac{AB}{3})$

= $\frac{h}{2}\times (DC+\frac{DC}{3})$

= $\frac{h}{2}\times \left(\frac{4DC}{3}\right)$

=$\frac{2h\times DC}{3}$

Now, we know that $h\times DC=$ area of parallelogram ABCD

Hence, area of trapezium DPQC = $\frac{2}{3}\times$ area of parallelogram ABCD

Thus,$\frac{AreaofDPQC}{AreaofABCD}=\frac{2}{3}$